The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is 4/3πr3. The surface area is 4πr2. Set up the differential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted?

Accepted Solution

Answer:According to the passage, we have the next equation:[tex]\frac{dV}{dt} = \frac{4}{3}\pi*3*r^{2}*\frac{dr}{dt} = K*(4\pi*r^{2} )[/tex]where "K" is a proportional constantLeaving at the end with the next equation:[tex]\frac{dr}{dt} = K[/tex]Integrating the equation, we have:[tex]r=K*t+C[/tex]where "C" is a constant Then, we have the 2 conditions for the problem:1) t=0 → r=10Replacing in the equation, we have C = 102) t=5 → r=8Replacing in the equation, we have K = -0.4Finally, the time which the snowball will be completely melted will be when r = 0. So replacing in the equation[tex]0=-0.4*t+10[/tex]t = 25 minutes